"Question 2 In the given figure sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
Class 9 - Math - Triangles Page 132"
Answers
If Non common arms of two adjacent angles form a line, then these angles are called linear pair of angles.
Axiom- 1
If a ray stands on a line, then the sum of two adjacent angles so formed is 180°i.e, the sum of the linear pair is 180°.
Axiom-2
If the sum of two adjacent angles is 180° then the two non common arms of the angles form a line.
The two axioms given above together are called the linear pair axioms.
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Use the Inequality that the side opposite to the larger angle is longer, to show the inequality.
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Given: ∆ABC is a Triangle &
∠PBC < ∠QCB
To Prove:
AC>AB
Proof:
∠ABC + ∠PBC = 180° ( by linear pair axiom)
⇒ ∠ABC = 180° – ∠PBC
also,
∠ACB + ∠QCB = 180° (by Linear Pair axiom)
⇒ ∠ACB = 180° – ∠QCB
Since,
∠PBC < ∠QCB therefore, ∠ABC > ∠ACB
Here, the side opposite to the larger angle is longer.
Hence, AC > AB
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Hope this will help you.....
Linear pair of angles:
If Non common arms of two adjacent angles form a line, then these angles are called linear pair of angles.
Axiom- 1
If a ray stands on a line, then the sum of two adjacent angles so formed is 180°i.e, the sum of the linear pair is 180°.
Axiom-2
If the sum of two adjacent angles is 180° then the two non common arms of the angles form a line.
The two axioms given above together are called the linear pair axioms.
____________________________________
Use the Inequality that the side opposite to the larger angle is longer, to show the inequality.
____________________________________
Given: ∆ABC is a Triangle &
∠PBC < ∠QCB
To Prove:
AC>AB
Proof:
∠ABC + ∠PBC = 180° ( by linear pair axiom)
⇒ ∠ABC = 180° – ∠PBC
also,
∠ACB + ∠QCB = 180° (by Linear Pair axiom)
⇒ ∠ACB = 180° – ∠QCB
Since,
∠PBC < ∠QCB therefore, ∠ABC > ∠ACB
Here, the side opposite to the larger angle is longer.
Hence, AC > AB